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Otrada [13]
2 years ago
6

If math were a weather, what type of weather would it be and why?

Mathematics
1 answer:
drek231 [11]2 years ago
8 0

Answer: a thunderstorm

Step-by-step explanation:

B/c there is more than one way to do math but if you don’t do it a specific way at a certain time then it becomes a problem.

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How do I solve this?
Artemon [7]

a^2 - b^2 = (a + b)(a - b)

In this case

49x^4y^2 - 4z^2

= (7x^2y)^2 - (2z)^2

= (7x^2y + 2z)(7x^2y - 2z)

Factors are (7x^2y + 2z) and (7x^2y - 2z)

Answer:

D) 7x^2y + 2z

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1. What kind of correlation is shown? (see #1 above)
klemol [59]
I this it would be constant correlation
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You are on top of a 1200 ft building. As
katrin [286]

Answer:

I parked my car 120 feet from the building

5 0
2 years ago
A gear with a diameter of 12.0 inches makes 35 revolutions every three minutes. Find the linear and angular velocities of a poin
alina1380 [7]

Answer:

Angular velocity: \omega = \frac{7\pi}{18} \,\frac{rad}{s} (1.222\,\frac{rad}{s})

Linear velocity: v = \frac{14\pi}{3}\,\frac{m}{s} (14.661\,\frac{m}{s})

Step-by-step explanation:

The gear experiments a pure rotation with axis passing through its center, the angular (\omega), in radians per second, and linear velocities (v), in inches per second, of a point on the outer edge of the element are, respectively:

\omega = \frac{2\pi}{60}\cdot \dot n (1)

v = R\cdot \omega (2)

Where:

\dot n - Rotation rate, in revolutions per minute.

R - Radius of the gear, in inches.

If we know that \dot n = \frac{35}{3}\,\frac{rev}{min} and R = 12\,in, then the linear and angular velocities of the gear are, respectively:

\omega = \frac{2\pi}{60}\cdot \dot n

\omega = \frac{7\pi}{18} \,\frac{rad}{s} (1.222\,\frac{rad}{s})

v = R\cdot \omega

v = \frac{14\pi}{3}\,\frac{m}{s} (14.661\,\frac{m}{s})

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2 years ago
What kind of triangle is BCD?
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The triangle BCD is an acute triangle
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2 years ago
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